访问方法内部的变量,并允许使用segue将其传递给其他视图控制器 [英] Accessing a variable inside a method and allowing it to be passed using segue to other view controller

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本文介绍了访问方法内部的变量,并允许使用segue将其传递给其他视图控制器的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

func collectionView(_ collectionView: UICollectionView, cellForItemAt indexPath: IndexPath) -> UICollectionViewCell{
let cell = collectionView.dequeueReusableCell(withReuseIdentifier: cellIdentifier, for: indexPath) as! FeedCollectionViewCell
cell.postTextView.text = posts.reversed()[indexPath.row].caption
cell.postImageView.image = UIImage(named: "photoplaceholder.jpg")
cell.priceTextView.text = posts.reversed()[indexPath.row].price
cell.categoryTextView.text = posts.reversed()[indexPath.row].category
cell.usernameLabel.text = posts.reversed()[indexPath.row].username
cell.buttonEvents = {
    let storyboard = UIStoryboard(name: "Main" , bundle: nil)
    let chatViewController =
        storyboard.instantiateViewController(withIdentifier: "chat")
    self.present(chatViewController, animated: true,completion: nil)
    var receiverIDNumber = cell.usernameLabel.text
}

我想将receiverIDNumber传递给另一个viewcontroller,但是由于该变量位于方法内部,所以我无法执行此操作.

I want to pass receiverIDNumber to another viewcontroller,however i am not able do as the variable is inside the method.

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {

    let chatViewController = segue.destination as! ChatViewController


    chatViewController.receivedString = receiverIDNumber
}

推荐答案

您可以尝试一下,因为 prepareForSegue 仅在 performSegue 不存在的情况下触发>

You can try this as prepareForSegue is only triggered with performSegue not present

let chatViewController = storyboard.instantiateViewController(withIdentifier: "chat") as! ChatViewController
chatViewController.receivedString = cell.usernameLabel.text
self.present(chatViewController, animated: true,completion: nil)

这篇关于访问方法内部的变量,并允许使用segue将其传递给其他视图控制器的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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